Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

class Main {


  

  public static void main(String[] args) {
    Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    int N = in.nextInt();

    double[] p = new double[N];
    for (int i = 0; i < N; i++) {p[i] = in.nextDouble();}

    double[][] dp = new double[N +  1][N + 1];
    dp[0][0] = 1;
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j <= i; j++) {
        dp[i][j] = (j == 0 ? 0 : (dp[i - 1][j - 1] * p[i - 1])) + dp[i - 1][j] * (1 - p[i - 1]);
      }
    }

    double ans = 0;
    for (int i = N / 2 + 1; i <= N; i++) {
      ans += dp[N][i];
    }




    System.out.println(ans);
  }

}